We know that the binomial theorem and expansion extends to powers which are non-integers.
For integer powers the expansion can be proven easily as the expansion is finite. However what is the proof that the expansion also holds for fractional powers?
A simple an intuitive approach would be appreciated.
Heres a proof:
Let $f(a)$ denote the binomial expansion of $(1+x)^a$ and $f(b)$ denote the same for $(1+x)^b$. Where
$(1+x)^a = 1 + ax.....$
$(1+x)^b = 1 + bx....$
On multiplying the two binomial expansions together....the product will be another series in ascending powers of $x$ and will remain unaltered irrespective of $a$ and $b$.
To determine this invariable form of the product we may give to $a$ and $b$ positive integral values for convenience.
Then
$f(a)×f(b)=(1+x)^{(a+b)}$
But when a and b are positive integers.... the expansion is
$1 + (a+b)x +.....$
This then is the form of the product of $f(a)×f(b)$ in all cases, whatever the values of a and b be; and in agreement with our previous notation it may be denoted by $f(a+b)$; therefore for all values of $a$ and $b$.
$f(a)×f(b)=f(a+b)$
Also
$f(a)×f(b)×f(p) = f(a+b+p)...$
Therefore
$f(a)×f(b)×f(p)... \text{to k factors} = f(a+b+p.... \text{to k terms}).$
Let each of these quantities a,b,p,.... be equal to $(c/k)$, where c and k are positive integers.
Hence
$f(c/k)^k = f(c)$
But since c is a positive integer,
$f(c)=(1+x)^c$
$(1+x)^c = f(c/k)^k$
Therefore
$(1+x)^{c/k} = f(c/k)$
And
$f(c/k) = 1 + (c/k)x ....$
Hence we get
$(1+x)^{c/k} = 1 + (c/k)x .....$
This proves the binomial theorem for any positive fractional index.